Two different groups in a statistics class are conducting a survey to estimate the mean number of units students at their college are enrolled in. The population mean and standard deviation are unknown. The sample from the first group survey has 49 data values. The sample from the second group survey has 81 data values. For each sample, the groups construct a 90% confidence interval to estimate the population mean. Which confidence interval will have greater precision (smaller width) for estimating the population mean?
a. The confidence interval based on the sample of 49 data values will be more precise.
b. Both confidence intervals will have the same precision.
c. The confidence interval based on the sample of 81 data values will be more precise.

idk

step-by-step explanation:

company a

step-by-step explanation:

using the linear regression models , it is found that company a   pays about $14 more. but using the same regression (linear) company b pays about $10 more.

so company a pays more than b

step-by-step explanation:

se a cavasery6

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: